Question: $f\,^{\prime}(x)=-3x^2+2x+4$ and $f(4)=-20$. $f(2) = $
Finding $f(x)$ We have $f'(x)=-3x^2+2x+4$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ & = \int (-3x^2+2x+4)\,dx \\\\ & = {-x^3+x^2+4x} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(4)=-20$. Here's what we get when we plug in $4$ : $\begin{aligned}f(4)&={-(4)^3+(4)^2+4(4)} {+ C}\\\\ &={-32} {+ C} \end{aligned}$ We are given that this must equal $-20$ : $-20 = {-32} {+ C}$ Solving the equation gives us ${C=12}$. Finding $f(2)$ Now, we have that $f(x)={-x^3+x^2+4x} {+12}$. Let's find $f(2)$ by plugging in $2$ : $\begin{aligned}f(2)&=-(2)^3+(2)^2+4(2) + 12\\\\ &=16 \end{aligned}$ The answer $f(2) = 16$